WAEC SSCE - Physics - 2001 (Essay)

Détails de la réponse

For the vertical motion of the projectile, the initial vertical velocity is \(u\sin\theta\) and the acceleration due to gravity is \(-g\).

Using

\[ h=(u\sin\theta)t-\frac{1}{2}gt^{2} \]

At maximum height,

\[ t=\frac{u\sin\theta}{g} \]

Therefore,

\[ h_{\text{max}}=(u\sin\theta)\left(\frac{u\sin\theta}{g}\right) -\frac{1}{2}g\left(\frac{u\sin\theta}{g}\right)^2 \] \[ h_{\text{max}}=\frac{u^2\sin^2\theta}{g} -\frac{1}{2}g\left(\frac{u^2\sin^2\theta}{g^2}\right) \] \[ h_{\text{max}}=\frac{u^2\sin^2\theta}{g} -\frac{u^2\sin^2\theta}{2g} \] \[ h_{\text{max}}=\frac{u^2\sin^2\theta}{2g} \]

Hence, the greatest height reached is

\[ \boxed{h_{\text{max}}=\frac{u^2\sin^2\theta}{2g}} \]

Détails de la réponse

(a) A beam of polarised light is light in which the vibrations are restricted to one direction or one plane only, perpendicular to the direction in which the light travels.

(b) As illustrated in the diagram above, sunlight is normally unpolarised. Its vibrations occur in many different planes perpendicular to the direction of travel.

When the sunlight falls on a horizontal surface such as a road, water surface or glass, the reflected light is largely polarised in a particular plane. This reflected polarised light produces glare.

A polaroid spectacle has a polaroid filter whose transmission axis is arranged vertically. The filter allows only the vertical component of the light vibrations to pass through it and absorbs the horizontal component, which is mainly responsible for glare reflected from horizontal surfaces.

• Unpolarised sunlight enters the spectacle lens.
• The polaroid lens acts as an analyser.
• Only vibrations parallel to the transmission axis of the lens emerge.
• The horizontally polarised reflected glare is greatly reduced or stopped.
• The wearer therefore sees more clearly and comfortably, especially when driving or viewing water.

Thus, polaroid spectacles reduce glare by selectively transmitting light vibrating in one plane only.

Détails de la réponse

(a) Stress and strain

Stress is the force acting per unit cross-sectional area of the material:

\[ \text{stress} = \frac{\text{force}}{\text{area}} \quad (\text{unit: N m}^{-2}\ \text{or Pa}) \]

Strain is the extension produced per unit original length of the material:

\[ \text{strain} = \frac{\text{extension}}{\text{original length}} \quad (\text{no unit, it is a ratio}) \]

Thus stress measures the intensity of the applied load while strain measures the relative deformation it produces.

(b) Work done on the cord

For an elastic cord obeying Hooke's law, the work done equals the elastic potential energy stored, given by the area under the force-extension graph:

\[ W = \tfrac{1}{2}Fe \]

With \(F = 40\,\text{N}\) and \(e = 5\,\text{cm} = 0.05\,\text{m}\):

\[ W = \tfrac{1}{2}(40)(0.05) = 1.0\,\text{J} \]

Détails de la réponse

(a) Transmutation equation

Nitrogen-14 bombarded by a neutron emits an alpha particle:

\[ {}^{14}_{7}\text{N} + {}^{1}_{0}\text{n} \;\rightarrow\; {}^{4}_{2}\text{He} + {}^{11}_{5}\text{B} \]

(Mass numbers: \(14+1 = 4+11\); atomic numbers: \(7+0 = 2+5\). The residual nucleus is boron-11.)

(b)(i) A radioactive atom has an unstable nucleus that spontaneously disintegrates, emitting radiation (alpha, beta or gamma), whereas a stable atom has a nucleus that does not decay.

(b)(ii) Half life is the time taken for half the atoms (or nuclei) in a given sample of a radioactive material to decay.

(b)(iii) \(105\,\text{days} = 3\times 35\,\text{days} = 3\) half lives. Fraction remaining:

\[ \left(\tfrac{1}{2}\right)^3 = \frac{1}{8} \]

(c) Photoelectric calculations (\(\lambda = 5.00\times10^{-7}\,\text{m}\), \(W_0 = 1.90\,\text{eV}\)):

(i) Photon energy:

\[ E = \frac{hc}{\lambda} = \frac{(6.6\times10^{-34})(3.0\times10^{8})}{5.0\times10^{-7}} = 3.96\times10^{-19}\,\text{J} \ (\approx 2.48\,\text{eV}) \]

(ii) Maximum kinetic energy: work function \(W_0 = 1.90\times1.6\times10^{-19} = 3.04\times10^{-19}\,\text{J}\).

\[ E_k = E - W_0 = 3.96\times10^{-19} - 3.04\times10^{-19} = 9.2\times10^{-20}\,\text{J} \ (0.575\,\text{eV}) \]

(iii) Stopping potential:

\[ V_s = \frac{E_k}{e} = \frac{9.2\times10^{-20}}{1.6\times10^{-19}} = 0.575\,\text{V} \]

Détails de la réponse

Given: maximum height \(h = 50\ \text{m}\), \(g = 10\ \text{m s}^{-2}\), and velocity at the top \(= 0\).

(ii) Speed of throw (find first): using \(v^{2} = u^{2} - 2gh\) with \(v = 0\):

\[ 0 = u^{2} - 2(10)(50) \Rightarrow u^{2} = 1000 \] \[ u = \sqrt{1000} = 31.6\ \text{m s}^{-1} \]

(i) Time to reach maximum height: using \(v = u - gt\) with \(v = 0\):

\[ t = \frac{u}{g} = \frac{31.6}{10} = 3.16\ \text{s} \]

The ball takes about 3.16 s to reach the top, having been thrown at about 31.6 m s\(^{-1}\).

Détails de la réponse

(a) How a gas can be made to conduct electricity

At ordinary pressure a gas is an insulator. It can be made to conduct by ionising it, that is by producing charged particles within it. This is done by applying a very high potential difference across the gas while it is at low pressure (as in a discharge tube), or by heating it strongly, or by exposing it to X-rays, ultraviolet or radioactive radiation. The high field/energy strips electrons from some of the gas atoms, producing free electrons and positive ions; under the applied field these charges drift to the electrodes and so a current flows.

(b) Electric charge carriers in gases

• Positive ions.
• Electrons (together with negative ions).

Détails de la réponse

(a)(i) An echo is a reflected sound wave heard distinctly after the original sound, produced when sound is reflected from a hard, distant surface back to the listener.

(a)(ii) Two useful applications:

• Depth sounding (echo sounding/SONAR) to measure the depth of the sea or locate shoals of fish and submarines.
• Detecting flaws in metals and, in medicine, ultrasonic scanning of the body.

(a)(iii) The walls, floors and ceilings of a recording studio are heavily padded with soft, porous material to absorb sound and prevent reflection (echoes and reverberation), so that only clear, direct sound is recorded.

(b)(i) Timbre (quality) is the property of a musical note that lets the ear distinguish two notes of the same pitch and loudness from different sources; it depends on the number and relative strength of the overtones present. Overtones are the higher frequencies (above the fundamental) that accompany a note.

(b)(ii) Resonance is the condition in which a body is set into vibration with large amplitude by a periodic force whose frequency equals the natural frequency of the body.

(c) Velocity of the ship

Distance of the cliff at the first sounding \(= \tfrac{1}{2}(340)(12) = 2040\,\text{m}\).

Distance at the second sounding \(= \tfrac{1}{2}(340)(8) = 1360\,\text{m}\).

Distance travelled by the ship between the two soundings:

\[ 2040 - 1360 = 680\,\text{m} \]

Time between soundings \(= 2.1\,\text{min} = 126\,\text{s}\). Hence:

\[ v = \frac{680}{126} \approx 5.4\,\text{ms}^{-1} \]

The ship approaches the cliff at about \(5.4\,\text{ms}^{-1}\).

Détails de la réponse

(a) Capacitance is defined by \(C=\dfrac{Q}{V}\). Therefore, a capacitance of \(5\,\mu\text{F}\) means that the capacitor stores a charge of \(5\,\mu\text{C}\) on each plate when the potential difference across it is \(1\,\text{V}\).

(b)(i) For a parallel-plate capacitor, capacitance depends on:

• the common area of overlap of the plates, \(A\);
• the separation between the plates, \(d\);
• the permittivity, \(\varepsilon\), of the dielectric between the plates.

(b)(ii) The relationship is

\[ C=\frac{\varepsilon A}{d}. \]

• \(C\) is directly proportional to plate area: \(C\propto A\).
• \(C\) is inversely proportional to plate separation: \(C\propto \dfrac{1}{d}\).
• \(C\) is directly proportional to the permittivity of the dielectric: \(C\propto\varepsilon\).

(c)(i) The capacitors are connected one after another in a single series path. The switch is shown open.

(c)(ii) For capacitors in series, the reciprocals of the capacitances are added:

\[ \frac{1}{C_{\text{effective}}} =\frac{1}{8}+\frac{1}{12}+\frac{1}{24}. \] \[ \frac{1}{C_{\text{effective}}} =\frac{3+2+1}{24} =\frac{6}{24} =\frac{1}{4}. \] \[ C_{\text{effective}}=4\,\mu\text{F}. \]

(c)(iii) In a series circuit, each capacitor has the same magnitude of charge. Thus:

\[ Q=C_{\text{effective}}V =(4\,\mu\text{F})(90\,\text{V}) =360\,\mu\text{C}. \]

Therefore, the charge on each of the \(8\,\mu\text{F}\), \(12\,\mu\text{F}\), and \(24\,\mu\text{F}\) capacitors is \(360\,\mu\text{C}\).

(c)(iv) The potential difference across the \(8\,\mu\text{F}\) capacitor is:

\[ V=\frac{Q}{C} =\frac{360\,\mu\text{C}}{8\,\mu\text{F}} =45\,\text{V}. \]

The p.d. across the \(8\,\mu\text{F}\) capacitor is therefore \(45\,\text{V}\). In a series combination, the smallest capacitance has the largest potential difference because \(V=\dfrac{Q}{C}\) and the charge is the same on all capacitors.

Détails de la réponse

The fall in e.m.f. with use is caused by two defects of the simple voltaic cell:

• Polarisation. As the cell works, hydrogen gas is liberated and clings to the copper (positive) plate, forming an insulating layer. This layer increases the internal resistance and sets up a back e.m.f., so the useful e.m.f. falls.
• Local action. Impurities (such as carbon and iron) in the commercial zinc plate form tiny local cells with the zinc, so the zinc is eaten away even when no current is being drawn, weakening the cell.

Corrections:

• Polarisation is removed by adding a depolariser (an oxidising agent such as manganese(IV) oxide or potassium dichromate) which oxidises the hydrogen to water.
• Local action is prevented by amalgamating the zinc plate, i.e. coating it with mercury so that the surface presents a clean layer of pure zinc only.

Détails de la réponse

(a) What is electrolysis?

Electrolysis is the chemical decomposition of an electrolyte (a molten ionic compound or an aqueous solution of one) caused by the passage of a direct electric current through it. The current makes the positive ions (cations) move to the cathode and the negative ions (anions) move to the anode, where they are discharged and new substances are set free.

(b) Mass of copper deposited

This uses Faraday's first law of electrolysis, which states that the mass \(m\) of a substance liberated at an electrode is proportional to the quantity of charge \(Q\) passed, so that \(m = zQ = zIt\), where \(z\) is the electrochemical equivalent.

Given: current \(I = 2\ \text{A}\); time \(t = 5\ \text{min} = 5 \times 60 = 300\ \text{s}\); electrochemical equivalent \(z = 3.27 \times 10^{-7}\ \text{kg C}^{-1}\).

Working:

\[ m = zIt = (3.27 \times 10^{-7}) \times 2 \times 300 \] \[ m = (3.27 \times 10^{-7}) \times 600 \] \[ m = 1.962 \times 10^{-4}\ \text{kg} \]

The mass of copper deposited is \(1.962 \times 10^{-4}\ \text{kg}\), which is about \(0.196\ \text{g}\).

Key point: the time must be converted from minutes to seconds before substituting, because the ampere is one coulomb per second; forgetting to change 5 min into 300 s is the most common error and would give an answer 60 times too small.

Détails de la réponse

Kinetic energy of the electron from its de Broglie wavelength

The de Broglie relation links wavelength to momentum:

\[ \lambda = \frac{h}{mv} \quad\Rightarrow\quad v = \frac{h}{m\lambda} \]

Substituting \(h = 6.6\times10^{-34}\,\text{Js}\), \(m = 9.0\times10^{-31}\,\text{kg}\), \(\lambda = 1.0\times10^{-10}\,\text{m}\):

\[ v = \frac{6.6\times10^{-34}}{(9.0\times10^{-31})(1.0\times10^{-10})} = \frac{6.6\times10^{-34}}{9.0\times10^{-41}} = 7.33\times10^{6}\,\text{ms}^{-1} \]

Then the kinetic energy is:

\[ E_k = \tfrac{1}{2}mv^2 = \tfrac{1}{2}(9.0\times10^{-31})(7.33\times10^{6})^2 \] \[ E_k = \tfrac{1}{2}(9.0\times10^{-31})(5.38\times10^{13}) \approx 2.42\times10^{-17}\,\text{J} \]

Equivalently, using \(E_k = \dfrac{h^2}{2m\lambda^2}\) gives the same result, \(E_k \approx 2.4\times10^{-17}\,\text{J}\).

Détails de la réponse

The graph is a straight line of the maximum kinetic energy \(E\) of the emitted photoelectrons (vertical axis, in joules) against the frequency \(f\) of the incident radiation (horizontal axis, in hertz). The line crosses the frequency axis at the point \(K\) and, when produced backwards, cuts the energy axis below the origin at the point \(C\). The graph is a plot of Einstein's photoelectric equation,

\[ E = hf - hf_0 \]

where \(h\) is Planck's constant and \(f_0\) is the threshold frequency of the metal. Comparing this with the equation of a straight line \(y = mx + c\): \(m = h\) is the gradient and \(c = -hf_0\) is the intercept on the energy axis.

(i) The slope of the graph

Comparing \(E = hf - hf_0\) with \(y = mx + c\), the gradient is

\[ \text{slope} = \frac{\Delta E}{\Delta f} = h \]

So the slope represents Planck's constant, \(h\) \(\left(\approx 6.6\times10^{-34}\,\text{J s}\right)\). It has the same value for every metal.

(ii) The intercept OC

OC is the intercept on the energy axis, obtained when \(f = 0\). From the equation, \(E = -hf_0\), so this intercept is negative and equal in magnitude to the work function \(\varphi\) of the metal:

\[ OC = -hf_0 = -\varphi \]

Thus \(|OC|\) is the minimum energy needed to release an electron from the surface of the metal, i.e. the work function.

(iii) The intercept OK

OK is the intercept on the frequency axis, obtained when the photoelectron energy \(E = 0\). Setting \(0 = hf - hf_0\) gives \(f = f_0\). Therefore OK represents the threshold (cut-off) frequency \(f_0\), the minimum frequency of the incident radiation below which no electrons are emitted from the metal.

Détails de la réponse

(a) Two properties of cathode rays

• They travel in straight lines from the cathode and cast sharp shadows of objects placed in their path.
• They are streams of negatively charged particles (electrons), so they are deflected by both electric and magnetic fields (towards the positive plate).

Other acceptable properties: they carry energy and momentum, they produce heat and fluorescence on striking a screen, and they can produce X-rays on hitting a metal target.

(b) Increasing the intensity and energy of cathode rays

Intensity (number of electrons per second) is increased by raising the filament (heater) current. A hotter cathode emits more electrons by thermionic emission, so the beam current and hence the brightness increase.

Energy (speed of each electron) is increased by raising the accelerating potential difference between the cathode and the anode. Each electron gains kinetic energy \(eV\) as it is accelerated, so a larger anode voltage gives faster, more energetic electrons.

Détails de la réponse

(a) Gain in the long jump

The horizontal range of a projectile is \(R = \dfrac{v^2\sin 2\theta}{g}\). With \(v = 8\,\text{ms}^{-1}\) and \(g = 10\,\text{ms}^{-2}\), \(\dfrac{v^2}{g} = \dfrac{64}{10} = 6.4\,\text{m}\).

First attempt (\(\theta = 30^\circ\)): \(R_1 = 6.4\sin 60^\circ = 6.4(0.8660) = 5.5424\,\text{m}\).

Second attempt (\(\theta = 45^\circ\)): \(R_2 = 6.4\sin 90^\circ = 6.4(1) = 6.4000\,\text{m}\).

\[ R_2 - R_1 = 6.4000 - 5.5424 = 0.8576\,\text{m} \quad\text{(shown)} \]

(b)(i) Hooke's law: Provided the elastic limit is not exceeded, the extension of an elastic material is directly proportional to the applied force (load).

(b)(ii) Experiment to verify Hooke's law: A spiral spring is clamped vertically with a pointer against a metre rule. The unstretched pointer reading is noted. Known masses are added one at a time; for each load the new pointer reading is taken and the extension found by subtraction. A graph of load against extension is plotted; a straight line through the origin verifies that extension is proportional to load, confirming Hooke's law.

(b)(iii) Two precautions:

• Do not exceed the elastic limit of the spring (avoid overloading so it returns to its original length).
• Take readings with the eye level to the pointer to avoid parallax error; allow the spring to settle before reading.

(c) Mass of the scale pan

Extension is proportional to the total load (pan mass \(m\) plus object). Natural length \(= 20.00\,\text{cm}\).

With 40 g: length \(21.80\,\text{cm}\), extension \(e_1 = 1.80\,\text{cm}\), load \(=(m+40)\).

With 60 g: length \(22.05\,\text{cm}\), extension \(e_2 = 2.05\,\text{cm}\), load \(=(m+60)\).

\[ \frac{m+40}{1.80} = \frac{m+60}{2.05} \] \[ 2.05(m+40) = 1.80(m+60) \] \[ 2.05m + 82 = 1.80m + 108 \;\Rightarrow\; 0.25m = 26 \;\Rightarrow\; m = 104\,\text{g} \]

The mass of the scale pan is \(104\,\text{g}\ (0.104\,\text{kg})\).

Détails de la réponse

(a) Specific heat capacity is the quantity of heat required to raise the temperature of unit mass (1 kg) of a substance by one kelvin (1 K). Its unit is \(\text{J kg}^{-1}\,\text{K}^{-1}\).

(b)(i) Experiment to determine the specific heat capacity of copper (method of mixtures)

Procedure:

• Weigh the empty copper calorimeter together with its stirrer; record the mass \(m_c\).
• Half-fill the calorimeter with water, weigh again, and obtain the mass of water \(m_w\) by subtraction. Note the initial temperature of the water and calorimeter, \(\theta_1\).
• Weigh the copper ball; record its mass \(m_b\). Suspend it by a thread inside a beaker of water and heat until the water boils steadily. Record the steady temperature of the ball, \(\theta\) (about \(100^\circ\text{C}\)).
• Shake off the adhering water and quickly transfer the hot ball into the calorimeter. Stir gently and record the highest steady temperature of the mixture, \(\theta_2\).

Since heat losses are negligible, heat lost by the copper ball = heat gained by the water + heat gained by the calorimeter and stirrer:

\[ m_b c_b(\theta-\theta_2) = m_w c_w(\theta_2-\theta_1) + m_c c_b(\theta_2-\theta_1) \]

Making \(c_b\), the specific heat capacity of copper, the subject:

\[ c_b = \frac{m_w c_w(\theta_2-\theta_1)}{m_b(\theta-\theta_2) - m_c(\theta_2-\theta_1)} \]

(b)(ii) Two precautions

• Transfer the hot copper ball to the calorimeter quickly and shake off any adhering water first, to minimise heat loss to the surroundings.
• Lag (insulate) the calorimeter and stir the water gently before reading the highest steady temperature.

(c) Final steady temperature of the mixture

Data: copper block \(m_b=24\text{ g}=0.024\text{ kg}\) at \(230^\circ\text{C}\); calorimeter \(m_c=60\text{ g}=0.060\text{ kg}\); water \(m_w=54\text{ g}=0.054\text{ kg}\) at \(31^\circ\text{C}\); \(c_w=4200\ \text{J kg}^{-1}\text{K}^{-1}\); \(c_b=400\ \text{J kg}^{-1}\text{K}^{-1}\). Let the final steady temperature be \(\theta\).

Heat lost by copper block = heat gained by water + heat gained by copper calorimeter:

\[ (0.024)(400)(230-\theta) = (0.054)(4200)(\theta-31) + (0.060)(400)(\theta-31) \] \[ 9.6(230-\theta) = (226.8 + 24)(\theta-31) = 250.8(\theta-31) \] \[ 2208 - 9.6\theta = 250.8\theta - 7774.8 \] \[ 2208 + 7774.8 = 250.8\theta + 9.6\theta \] \[ 9982.8 = 260.4\,\theta \] \[ \theta = \frac{9982.8}{260.4} = 38.34^\circ\text{C} \]

The final steady temperature of the mixture is \(\theta \approx 38.3^\circ\text{C}\).